Math

QuestionDetermine the limiting reagent and maximum moles of silver chloride from 0.581 moles of silver nitrate and 0.195 moles of copper(II) chloride.

Studdy Solution

STEP 1

Assumptions1. The reaction is balanced and proceeds to completion. . The reaction is as follows AgNO3(aq) + CuCl(s) -> AgCl(s) + Cu(NO3)(aq)
3. The initial amount of silver nitrate (AgNO3) is0.581 moles.
4. The initial amount of copper(II) chloride (CuCl) is0.195 moles.
5. The stoichiometry of the reaction is1111, which means for every mole of AgNO3 reacted, one mole of CuCl is consumed, and one mole of AgCl and one mole of Cu(NO3) are produced.

STEP 2

To find the limiting reagent, we need to compare the mole ratio of the reactants with the stoichiometry of the reaction. The limiting reagent is the reactant that is completely consumed first.

STEP 3

The stoichiometry of the reaction is11 for AgNO3 and CuCl2. This means that for every mole of AgNO3, we need one mole of CuCl2.

STEP 4

Calculate the available mole ratios for AgNO3 and CuCl2.
RatioAgNO3=0.581moles1=0.581Ratio_{AgNO3} = \frac{0.581\, moles}{1} =0.581RatioCuCl2=0.195moles1=0.195Ratio_{CuCl2} = \frac{0.195\, moles}{1} =0.195

STEP 5

Compare the calculated ratios. The reactant with the smaller ratio is the limiting reagent.
RatioAgNO3>RatioCuCl2Ratio_{AgNO3} > Ratio_{CuCl2}So, CuCl2 is the limiting reagent.

STEP 6

To find the maximum amount of silver chloride (AgCl) that can be produced, we need to use the stoichiometry of the reaction and the amount of the limiting reagent.

STEP 7

The stoichiometry of the reaction is1111, which means for every mole of CuCl2 consumed, one mole of AgCl is produced.

STEP 8

Calculate the maximum amount of AgCl that can be produced using the amount of the limiting reagent (CuCl2).
AmountAgCl=AmountCuCl2=0.195molesAmount_{AgCl} = Amount_{CuCl2} =0.195\, molesThe limiting reagent is CuCl2 and the maximum amount of AgCl that can be produced is0.195 moles.

Was this helpful?

Studdy solves anything!

banner

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

ParentsInfluencer programContactPolicyTerms
TwitterInstagramFacebookTikTokDiscord